We consider termination of the LCTRS with only rule scheme Calc:

  Signature: f   :: Int -> Int -> Int
             h   :: Int -> Int -> Int
             if1 :: Bool -> Int -> Int -> Int
             if2 :: Bool -> Int -> Int -> Int

  Rules: f(x, y) -> if1(x > y, x, y) | x = x /\ y = y
         h(x, y) -> if2(x > y, x, y) | x = x /\ y = y
         if1(true, x, y) -> h(x, y) | x = x /\ y = y
         if1(false, x, y) -> 0 | x = x /\ y = y
         if2(true, x, y) -> 0 | x = x /\ y = y
         if2(false, x, y) -> f(x, y) | x = x /\ y = y

The system is accessible function passing by a sort ordering that equates all sorts.

We start by computing the initial DP problem D1 = (P1, R, i, c), where:

  P1. (1) f#(x, y) => if1#(x > y, x, y) | x = x /\ y = y
      (2) h#(x, y) => if2#(x > y, x, y) | x = x /\ y = y
      (3) if1#(true, x, y) => h#(x, y) | x = x /\ y = y
      (4) if2#(false, x, y) => f#(x, y) | x = x /\ y = y

***** We apply the Chaining Processor Processor on D1 = (P1, R, i, c).

We chain DPs according to the following mapping:


  if2#(false, x, y) => if1#(x > y, x, y) | x = x /\ y = y /\ (x = x /\ y = y)  is obtained by chaining  if2#(false, x, y) => f#(x, y) | x = x /\ y = y and f#(x', y') => if1#(x' > y', x', y') | x' = x' /\ y' = y'


The following DPs were deleted:

if2#(false, x, y) => f#(x, y) | x = x /\ y = y

f#(x, y) => if1#(x > y, x, y) | x = x /\ y = y


By chaining, we added 1 DPs and removed 2 DPs.

Processor output: { D2 = (P2, R, i, c) }, where:

  P2. (1) h#(x, y) => if2#(x > y, x, y) | x = x /\ y = y
      (2) if1#(true, x, y) => h#(x, y) | x = x /\ y = y
      (3) if2#(false, x, y) => if1#(x > y, x, y) | x = x /\ y = y /\ (x = x /\ y = y)

***** We apply the Chaining Processor Processor on D2 = (P2, R, i, c).

We chain DPs according to the following mapping:


  if1#(true, x, y) => if2#(x > y, x, y) | x = x /\ y = y /\ (x = x /\ y = y)  is obtained by chaining  if1#(true, x, y) => h#(x, y) | x = x /\ y = y and h#(x', y') => if2#(x' > y', x', y') | x' = x' /\ y' = y'


The following DPs were deleted:

if1#(true, x, y) => h#(x, y) | x = x /\ y = y

h#(x, y) => if2#(x > y, x, y) | x = x /\ y = y


By chaining, we added 1 DPs and removed 2 DPs.

Processor output: { D3 = (P3, R, i, c) }, where:

  P3. (1) if2#(false, x, y) => if1#(x > y, x, y) | x = x /\ y = y /\ (x = x /\ y = y)
      (2) if1#(true, x, y) => if2#(x > y, x, y) | x = x /\ y = y /\ (x = x /\ y = y)

***** We apply the Usable Rules Processor on D3 = (P3, R, i, c).

We obtain 0 usable rules (out of 6 rules in the input problem).

Processor output: { D4 = (P3, {}, i, c) }.

***** No progress could be made on DP problem D4 = (P3, {}, i, c).